SOLUTION: if the number of bacteria in a culture is given by the following equation and the number of bacteria that exist after 1 hour is 825, how many exist after 10 hours? N(t)=195e^kt
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Question 173141: if the number of bacteria in a culture is given by the following equation and the number of bacteria that exist after 1 hour is 825, how many exist after 10 hours? N(t)=195e^kt
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
if the number of bacteria in a culture is given by the following equation and the number of bacteria that exist after 1 hour is 825, how many exist after 10 hours? N(t)=195e^kt
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N(1) = 195e^(k*1) = 825
e^k = 825/195
e^k = 4.230769
Take the natural log to find "k".
k = ln(4.230769)= 1.4423838...
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Therefore N(t) 195e^(1.4423838t)
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Then N(10) = 195e^(14.423838) = 358284370 bacteria
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Cheers,
Stan H.
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